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I am trying to prove the following:

(I) Let $f:X\rightarrow \mathbb{C}$ (i.e. complex space) and $f=u+iv$ where $u$ is the real part and $v$ is the imaginary part. If $f$ is measurable, then the magnitude of $f$ given as $|f|$ is measurable.

(II) If $f$ is measurable then the sign function of $f$ given as

$$sign(f)=\begin{cases}\frac{f}{|f|},&f\neq 0,\\0,&else,\end{cases}$$

is also measurable.

Thanks in advance

Deliasaghi
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Ahmad Abdulrahman
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  • Just prove that if $f:X\to \mathbb{C}$ is measurable and $\varphi : \mathbb{C}\to \mathbb{C}$ continuous, then $\varphi\circ f$ is measurable. – Prahlad Vaidyanathan Sep 23 '15 at 01:31
  • I am not sure how this would prove this. I was thinking of using the definition of the magnituge |f|=(f^+) + (f^-) and take it from there but not sure how? – Ahmad Abdulrahman Sep 23 '15 at 01:36
  • What is your definition of measurability of a function? – Prahlad Vaidyanathan Sep 23 '15 at 01:37
  • f is measurable if {x:f(x)>a} is in the sigma algebra of the measure space for all real numbers a. – Ahmad Abdulrahman Sep 23 '15 at 01:39
  • Well, this is the set $f^{-1}((a,\infty))$. Can you prove that this is equivalent to saying that $f^{-1}(U)$ is measurable for any open set $U \subset \mathbb{C}$? From that, the statement I made above follows quite easily. – Prahlad Vaidyanathan Sep 23 '15 at 01:42
  • I think I need further explanation although I understand that essentially what I need to prove for part (i) is that $1/|f|=\phi$ is measurable and then the composition of $f$ and $\phi$ is measurable. And how about the first part? Thanks – Ahmad Abdulrahman Sep 23 '15 at 02:01
  • @AhmadAbdulrahman: Be careful: your definition of "measurable function" only applies to real-valued functions. If $f$ is complex-valued, $f(x)>a$ makes no sense. – Giuseppe Negro Sep 23 '15 at 03:29
  • @AhmadAbdulrahman: You just need to prove that $x\mapsto |x|$ is continuous. Why worry about $1/|f|$? – Prahlad Vaidyanathan Sep 23 '15 at 06:48

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For both parts, you can use the fact that composition of measurable functions gives a measurable function*. With that in mind, for part 1, you only have to show that the magnitude function $z \mapsto |z|$ from $\mathbb{C} \to \mathbb{R}$ is Borel measurable. This function is continuous, and is therefore Borel measurable. For part 2, we need to show that $sign$ is Borel measurable. Use the fact that we can write $sign$ in countably many (even finitely many) continuous pieces (with Borel domains).

*As per a comment, we need to be careful about $\sigma$-algebras a little bit. If we are considering $f \circ g$ with, e.g., $g:X \to \mathbb{R}$ and $f:\mathbb{R} \to \mathbb{R}$, we need to be talking about the same $\sigma$-algebra for the domain of $f$ and codomain of $g$. To say that $g$ is measurable usually means (as it does in your definition) that we are talking about the Borel $\sigma$-algebra on $\mathbb{R}$, so we should be making sure that $f$ is a Borel-measurable function, also, in that the inverse image of each $\{x : x > a\}$ is a Borel set. The approaches I outline do show measurability in this stronger sense.

  • What you say is not true: http://math.stackexchange.com/questions/283443/is-composition-of-measurable-functions-measurable – Prahlad Vaidyanathan Sep 23 '15 at 05:34
  • @PrahladVaidyanathan Added caveat to clarify we need to show that the outer function in the composition is Borel-measurable. –  Sep 23 '15 at 10:30